Usually Diazo coupling occurs at para-position unless the para-position is occupied, in which case coupling occurs at ortho-position.

While solving questions I found in both the cases as shown, the coupling occurred at ortho-position of the $\ce{-OH}$ group:

enter image description here

So in case of phenol, is it possible that in all cases the coupling will occur at ortho-position with respect to the $\ce{-OH}$ group in phenol, irrespective of whether the other group has stronger +M Effect than $\ce{-OH}$?

I am not sure but I think this is happening because of Hydrogen bonding as shown below. But in that case it should also happen with the $\ce{-NH2}$ group. Is this the reason or there is some other?

enter image description here

  • $\begingroup$ If there’s hydroxide ion there, then the phenol will be deprotonated. $\endgroup$ – orthocresol Mar 21 '19 at 20:17
  • $\begingroup$ @orthocresol True , the Books Says Appreciable amount of Phenol will be converted in phenoxide ion . But Still the question and options all have -OH . $\endgroup$ – Chloritone_360 Mar 21 '19 at 20:42
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    $\begingroup$ Don't get so caught up about what the question writes... You put in a phenol into the reaction, it gets converted into phenoxide, that's all there is to it. Just because somebody drew a phenol on a piece of paper doesn't mean that it can't be deprotonated. $\endgroup$ – orthocresol Mar 21 '19 at 21:01
  • $\begingroup$ I'm not sure what you're trying suggest with that hydrogen bond. This is probably an exergonic reaction, in which case, the Hammond Postulate would suggest an early transition state where the form of the product has little impact on the rate of the reaction... $\endgroup$ – Zhe Mar 21 '19 at 23:26

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